What is the Pumping Lemma in Layman's terms?

Question

I saw this question, and was curious as to what the pumping lemma was (Wikipedia didn't help much).

I understand that it's basically a theoretical proof that must be true in order for a language to be in a certain class, but beyond that I don't really get it.

Anyone care to try to explain it at a fairly granular level in a way understandable by non mathematicians/comp sci doctorates?

Graniela

I hold the firm belief that there is no shortcut to mathematics/TCS: "laymen's terms" won't get you to understanding. That said, we have of course covered this extensively on Computer Science; see here, here and here. – Beatriz

Note that first-year students are routinely expected to understand the theorem and its proof, and apply it, so the request for something "understandable by non [...] doctorates" is easily fulfilled by looking into any formal languages textbook. – Beatriz

The pumping lemma is not a proof: as the name suggests, it's a lemma. – Mossberg

Answer 1

The pumping lemma is a simple proof to show that a language is not regular, meaning that a Finite State Machine cannot be built for it. The canonical example is the language (a^n)(b^n). This is the simple language which is just any number of as, followed by the same number of bs. So the strings

ab
aabb
aaabbb
aaaabbbb

etc. are in the language, but

aab
bab
aaabbbbbb

etc. are not.

It's simple enough to build a FSM for these examples:

This one will work all the way up to n=4. The problem is that our language didn't put any constraint on n, and Finite State Machines have to be, well, finite. No matter how many states I add to this machine, someone can give me an input where n equals the number of states plus one and my machine will fail. So if there can be a machine built to read this language, there must be a loop somewhere in there to keep the number of states finite. With these loops added:

all of the strings in our language will be accepted, but there is a problem. After the first four as, the machine loses count of how many as have been input because it stays in the same state. That means that after four, I can add as many as as I want to the string, without adding any bs, and still get the same return value. This means that the strings:

aaaa(a*)bbbb

with (a*) representing any number of as, will all be accepted by the machine even though they obviously aren't all in the language. In this context, we would say that the part of the string (a*) can be pumped. The fact that the Finite State Machine is finite and n is not bounded, guarantees that any machine which accepts all strings in the language MUST have this property. The machine must loop at some point, and at the point that it loops the language can be pumped. Therefore no Finite State Machine can be built for this language, and the language is not regular.

Remember that Regular Expressions and Finite State Machines are equivalent, then replace a and b with opening and closing Html tags which can be embedded within each other, and you can see why it is not possible to use regular expressions to parse Html

Karolkarola

Your second diagram is also incorrect in that it can produce baaaabbbb. – Corliss

@Corliss that is true, it could be fixed pretty simply by adding another accepting state but just for the sake of simplicity I'll leave it as is. – Karolkarola

Good answer but makes no mention that the pumping lemma can be used to prove that a language IS context free, not just disprove regularity – Penn

This does not even conclusively show that a^n b^n is non-regular, nor does it propose much of an intuition about the Pumping lemma. – Beatriz

@GraphicsNoob The pumping lemma is NOT a proof, it's a lemma, as the name suggests. A lemma is a proposition that has been proved. A lemma can be thought as a smaller not so important theorem, that is usually used for proving or showing other propositions or statements. I don't believe that an answer that starts saying that "the pumping lemma is a proof" has currently 114 up-votes, this is why questions and answers should be up-voted with a description or an explanation. – Mossberg

@Nuncameesquecideti or you can say that there is a mass who needs to understand things in simpler terms, why things like theoretical cs be a sport for math nerd only? – Conglobate

@GraphicsNoob is there a similar intuition for CFG? – Conglobate

Answer 2

It's a device intended to prove that a given language cannot be of a certain class.

Let's consider the language of balanced parentheses (meaning symbols '(' and ')', and including all strings that are balanced in the usual meaning, and none that aren't). We can use the pumping lemma to show this isn't regular.

(A language is a set of possible strings. A parser is some sort of mechanism we can use to see if a string is in the language, so it has to be able to tell the difference between a string in the language or a string outside the language. A language is "regular" (or "context-free" or "context-sensitive" or whatever) if there is a regular (or whatever) parser that can recognize it, distinguishing between strings in the language and strings not in the language.)

LFSR Consulting has provided a good description. We can draw a parser for a regular language as a finite collection of boxes and arrows, with the arrows representing characters and the boxes connecting them (acting as "states"). (If it's more complicated than that, it isn't a regular language.) If we can get a string longer than the number of boxes, it means we went through one box more than once. That means we had a loop, and we can go through the loop as many times as we want.

Therefore, for a regular language, if we can create an arbitrarily long string, we can divide it into xyz, where x is the characters we need to get to the start of the loop, y is the actual loop, and z is whatever we need to make the string valid after the loop. The important thing is that the total lengths of x and y are limited. After all, if the length is greater than the number of boxes, we've obviously gone through another box while doing this, and so there's a loop.

So, in our balanced language, we can start by writing any number of left parentheses. In particular, for any given parser, we can write more left parens than there are boxes, and so the parser can't tell how many left parens there are. Therefore, x is some amount of left parens, and this is fixed. y is also some number of left parens, and this can increase indefinitely. We can say that z is some number of right parens.

This means that we might have a string of 43 left parens and 43 right parens recognized by our parser, but the parser can't tell that from a string of 44 left parens and 43 right parens, which isn't in our language, so the parser can't parse our language.

Since any possible regular parser has a fixed number of boxes, we can always write more left parens than that, and by the pumping lemma we can then add more left parens in a way that the parser can't tell. Therefore, the balanced parenthesis language can't be parsed by a regular parser, and therefore isn't a regular expression.

Kollwitz

Excellent answer and read for those seeking to capture balanced strings with regular expressions. – Inerney

Answer 3

Its a difficult thing to get in layman's terms, but basically regular expressions should have a non-empty substring within it that can be repeated as many times as you wish while the entire new word remains valid for the language.

In practice, pumping lemmas are not sufficient to PROVE a language correct, but rather as a way to do a proof by contradiction and show a language does not fit in the class of languages (Regular or Context-Free) by showing the pumping lemma does not work for it.

Gonroff

What do you mean "not sufficient to PROVE a language correct"? By "correct" I guess you meant regular. Indeed, a regular language exhibits the pumping property, but if a language exhibits the pumping property it does not mean necessarily it's regular. On the other hand, if the language does not exhibit the pumping property, then we are sure it's not regular. Basically, the pumping property is necessary but not sufficient to show that a language is regular. – Mossberg

Answer 4

Basically, you have a definition of a language (like XML), which is a way to tell whether a given string of characters (a "word") is a member of that language or not.

The pumping lemma establishes a method by which you can pick a "word" from the language, and then apply some changes to it. The theorem states that if the language is regular, these changes should yield a "word" that is still from the same language. If the word you come up with isn't in the language, then the language could not have been regular in the first place.

Westlund

Answer 5

The simple pumping lemma is the one for regular languages, which are the sets of strings described by finite automata, among other things. The main characteristic of a finite automation is that it only has a finite amount of memory, described by its states.

Now suppose you have a string, which is recognized by a finite automaton, and which is long enough to "exceed" the memory of the automation, i.e. in which states must repeat. Then there is a substring where the state of the automaton at the beginning of the substring is the same as the state at the end of the substring. Since reading the substring doesn't change the state it may be removed or duplicated an arbitrary number of times, without the automaton being the wiser. So these modified strings must also be accepted.

There is also a somewhat more complicated pumping lemma for context-free languages, where you can remove/insert what may intuitively be viewed as matching parentheses at two places in the string.

Horeb

Your second paragraph is nice, but the first one is a little bit bad: "The simple pumping lemma is the one for regular languages". Is the one for regular languages to do what? Why do we need the pumping lemma? What's the relation ship between the pumping lemma and being a regular language? You should answer all these questions, IMO. – Mossberg

@starblue:Could you tell why If the language is ${a}$, the minimum pumping length is $2$; if the language is ${a^n:n∈ℕ}$, then the minimum pumping length is $1$.more at here:(math.stackexchange.com/questions/1508471/minimum-pumping-length/…). – Crews

Answer 6

By definition regular languages are those recognized by a finite state automaton. Think of it as a labyrinth : states are rooms, transitions are one-way corridors between rooms, there's an initial room, and an exit (final) room. As the name 'finite state automaton' says, there is a finite number of rooms. Each time you travel along a corridor, you jot down the letter written on its wall. A word can be recognized if you can find a path from the initial to the final room, going through corridors labelled with its letters, in the correct order.

The pumping lemma says that there is a maximum length (the pumping length) for which you can wander through the labyrinth without ever going back to a room through which you have gone before. The idea is that since there are only so many distinct rooms you can walk in, past a certain point, you have to either exit the labyrinth or cross over your tracks. If you manage to walk a longer path than this pumping length in the labyrinth, then you are taking a detour : you are inserting a(t least one) cycle in your path that could be removed (if you want your crossing of the labyrinth to recognize a smaller word) or repeated (pumped) indefinitely (allowing to recognize a super-long word).

There is a similar lemma for context-free languages. Those languages can be represented as word accepted by pushdown automata, which are finite state automata that can make use of a stack to decide which transitions to perform. Nonetheless, since there is stilla finite number of states, the intuition explained above carries over, even through the formal expression of the property may be slightly more complex.

Rb

@Looking for an answer like this.Could the initial and final room be the same?I'm stuck with this comment:If the language is ${a}$, the minimum pumping length is $2$; if the language is ${a^n:n∈N}$, then the minimum pumping length is $1$.Could you help me.more at here:(math.stackexchange.com/questions/1508471/minimum-pumping-length/…). – Crews

Answer 7

In laymans terms, I think you have it almost right. It's a proof technique (two actually) for proving that a language is NOT in a certain class.

Fer example, consider a regular language (regexp, automata, etc) with an infinite number of strings in it. At a certain point, as starblue said, you run out of memory because the string is too long for the automaton. This means that there has to be a chunk of the string that the automaton can't tell how many copies of it you have (you're in a loop). So, any number of copies of that substring in the middle of the string, and you still are in the language.

This means that if you have a language that does NOT have this property, ie, there is a sufficiently long string with NO substring that you can repeat any number of times and still be in the language, then the language isn't regular.

Aweinspiring

The last sentence, at least, is false. The language consisting of the string "a" is regular, but you can't pump it. If you can pump a string in a certain way, it isn't regular. For example, the language with symbols '(' and ')', made of all balanced expressions (and no unbalanced ones) isn't regular, and you prove that by pumping "()". – Kollwitz

@David, thanks, corrected last sentence. But I think you're wrong about balanced parens. I don't think you can prove parens isn't regular via pumping lemma. I think parens pumps. – Aweinspiring

Answer 8

For example, take this language L = aⁿbⁿ.

Now try to visualize finite automaton for the above language for some n's.

if n = 1, the string w = ab. Here we can make a finite automaton with out looping if n = 2, the string w = a²b². Here we can make a finite automaton with out looping

if n = p, the string w = a^pb^p. Essentially a finite automaton can be assumed with 3 stages. First stage, it takes a series of inputs and enter second stage. Similarly from stage 2 to stage 3. Let us call these stages as x, y and z.

There are some observations

1. Definitely x will contain 'a' and z will contain 'b'.
2. Now we have to be clear about y:
   • case a: y may contain 'a' only
   • case b: y may contain 'b' only
   • case c: y may contain a combination of 'a' and 'b'

So the finite automaton states for stage y should be able to take inputs 'a' and 'b' and also it should not take more a's and b's which cannot be countable.

1. If stage y is taking only one 'a' and one 'b', then there are two states required
2. If it is taking two 'a' and one 'b', three states are required with out loops and so on....

So the design of stage y is purely infinite. We can only make it finite by putting some loops and if we put loops, the finite automaton can accept languages beyond L = aⁿbⁿ. So for this language we can't construct a finite automaton. Hence it is not regular.

Gaekwar

Answer 9

This is not an explanation as such but it is simple. For a^n b^n our FSM should be built in such a way that b must know the number of a's already parsed and will accept the same n number of b's. A FSM can not simply do stuff like that.

Tippler